$H_m(\mathbb{R}^n)$ , the completion of $C_C^{\infty}(\mathbb{R}^n)$

Question

Theorem: Let $m$ be a positive integer. Then $H_m(\mathbb{R}^n)=\{ u \in D'(\mathbb{R}^n): D^{\alpha} u \in L^2(\mathbb{R}^n), |\alpha| \leq m\}$

$\to ||u||_{H_m}^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}<+\infty$

Furthermore, $H_m(\mathbb{R}^n)$ is the completion of $C_C^{\infty}(\mathbb{R}^n)$ ad for the metric $( \star) ||u||_m=\left( \sum_{|\alpha| \leq m} \int |D^{\alpha}u|^2\right)^{\frac{1}{2}} $

Proof: There are constants $c_m$ such that $$(1) \sum_{|\alpha| \leq m } |\xi^{\alpha}|^2 \leq (1+ |\xi|^2)^m \leq c_m \sum_{|\alpha| \leq m} |\xi^{\alpha}|^2$$

We have $\xi^{\alpha} \widehat{u}(\xi)=\widehat{D^{\alpha} u}$, so $D^{\alpha} u \in L^2(\mathbb{R}^n) \Rightarrow \xi^{\alpha} \widehat{u}(\xi) \in L^2(\mathbb{R}^n)$

From $(1)$ we have that the two metrics $(\star)$ and $||u||_m^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}$ are equivalent.

Claim: each Cauchy sequence $(\phi_j)_{j=1}^{\infty} \in C_C^{\infty}(\mathbb{R}^n)$ as for the metric $||\cdot||_m$ converges to an element of $H_m(\mathbb{R}^n)$.

We have that $\phi_j$ is Cauchy in $L^2(\mathbb{R}^n)$, $D^{\alpha}\phi_j$ is Cauchy in $L^2(\mathbb{R}^n)$.

$||\phi_j||_m^2=\sum_{|\alpha| \leq m} |D^{\alpha} \phi_j|^2 dx$

Let $u$ the $L^2$-limit of $\phi_j$ and $u^{\alpha}$ is the $L^2$-limit of $D^{\alpha}\phi_j$ , then

$$\int_{\mathbb{R}^n} \psi D^{\alpha} \phi_j=(-1)^{|\alpha|} \int_{\mathbb{R}^n} D^{\alpha} \psi \phi_j dx, \psi \in C_C^{\infty}(\mathbb{R}^n)$$

$$\Rightarrow \int_{\mathbb{R}^n} \psi u^{\alpha} dx=(-1)^{|\alpha|} \int D^{\alpha}{\psi} u dx \Rightarrow u^{\alpha}=D^{\alpha} u$$

Furthermore, each function $u \in H_m(\mathbb{R}^n)$ is the limit of a sequence $(\phi_j)_{j=1}^{\infty}$.

We define $u_{\epsilon}(x)= \int_{\mathbb{R}^n} \psi_{\epsilon}(x-y) u(y) dy$.

$$||u_{\epsilon}||_{L^2} \leq ||u||_{L^2}, ||D^{\alpha} u_{\epsilon}||_{L^2} \leq ||D^{\alpha} u||_{L^2}$$

$$||D^{\alpha} u_{\epsilon}- D^{\alpha} u||_{L^2(\mathbb{R}^n)} \to 0$$

We consider a cutoff function $\phi \in C_C^{\infty}(\mathbb{R}^n), 0 \leq \phi \leq 1$ and $\phi=1$ for $|x|<1$.

We define $g_{\epsilon}(x)=\phi( \epsilon x) u_{\epsilon}(x)$.

$D^{\alpha} g_{\epsilon}-D^{\alpha} u=D^{\alpha} (\phi(\epsilon x) u_{\epsilon }x)-D^{\alpha} u(x)= \phi(\epsilon x) D^{\alpha} u_{\epsilon}(x)- \phi(\epsilon x) D^{\alpha} u+ \phi(\epsilon x) D^{\alpha}u-D^{\alpha} u+ \epsilon(\dots)=\phi(\epsilon x) (D^{\alpha} u_{\epsilon}(x)-D^{\alpha}u)+(\phi(\epsilon x)-1) D^{\alpha}u+ \epsilon (\dots)$

$\Rightarrow D^{\alpha} g_{\epsilon} \overset{L^2}{\rightarrow} D^{\alpha}u$

Firts of all, how does the relation $(1)$ imply that the metrics $(\star)$ and $||u||_m^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}$ are eqivalent?

Secondly, in order to show that $H_m(\mathbb{R}^n)$ is the completion of $C_{C}^{\infty}(\mathbb{R}^n)$, do we need to show that each sequence of $C_C^{\infty}(\mathbb{R}^n)$ converges to an element of $H_m(\mathbb{R}^n)$ and that each element of $H_m(\mathbb{R}^n)$ is the limit of a sequence of $C_C^{\infty}(\mathbb{R}^n)$ ? If so, why?

Why are $\phi_j$ and $D^{\alpha} \phi_j$ Cauchy sequences in $L^2(\mathbb{R}^n)$ ?

In addition, why do we look for the difference $D^{\alpha}g_{\epsilon}-D^{\alpha}u $ ?

And how did we calculate $D^{\alpha}(\phi( \epsilon x) u_{\epsilon}(x))$ ?

Answer 1

The first question follows by definition, $H_m(\mathbb{R}^n)$ is the completion of $C_{c}^\infty(\mathbb{R}^n)$ iff (by definition) $H_m(\mathbb{R}^n) = \overline{C_{c}^\infty(\mathbb{R}^n)}^{|| \cdot ||_m}$, i.e. $\forall u \in H_m(\mathbb{R}^n)$ exists $\lbrace u_k \rbrace \subset C_{c}^\infty(\mathbb{R}^n)$ such that $u_k \rightarrow u$ in the $H_m$-norm. The second question follows because the test functions (with their derivatives) are uniformly limited. The third question follows by approximation theorem with regular functions, in this case consider the regularized function (convolution) of the weak derivative, that precisely approximates the weak derivative. The last question should follow from the Leibniz rule.

Note that the point where it checks that $u_\alpha = D^\alpha u$ follows by Schwartz inequality, in the sense

$\displaystyle \left \vert \int \psi [D^\alpha \phi_j - u_\alpha] dx \right \vert \leq ||\psi||_2 ||D^\alpha \phi_j - u_\alpha ||_2 \rightarrow 0 $

then

$\displaystyle \int \psi D^\alpha \phi_j dx = (-1)^{|\alpha|} \int \phi_j D^\alpha \psi dx \rightarrow (-1)^{|\alpha|} \int u_\alpha D^\alpha \psi dx$

by definition, $\forall \psi \in C^{\infty}_c(\mathbb{R}^n)$, this means that ($\star$) $u_\alpha = D^\alpha u$ is a weak derivative. The point ($\star$) follow by this lemma

"Let $f_n \in L^{1}_{loc}(\Omega)$ with $f_n$ admits weak derivative $g_n=D^\alpha f_n$. If $f_n \rightarrow f$ and $g_n \rightarrow g$ in $L^{1}_{loc}(\Omega)$ then $g=D^\alpha f$"

Proof. $\forall \phi \in C_{c}^\infty(\Omega)$ we have

$\displaystyle \int_{\Omega} g \phi dx = \lim_{n \rightarrow \infty} \int_{\Omega} g_n \phi dx = \lim_{n \rightarrow \infty} (-1)^{|\alpha|} \int_{\Omega} f_n D^\alpha \phi dx = (-1)^{|\alpha|} \int_{\Omega} f D^\alpha \phi dx$